Here \( \antisymmetric{A}{B} \ne 0 \), and \( \antisymmetric{A}{H} = \antisymmetric{B}{H} = 0 \). I don’t see a way for any exception to be constructed.

The problem

The concrete examples above give some intuition for solving the more abstract problem. Suppose that we are working in a basis that simulaneously diagonalizes operator \( A \) and the Hamiltonian \( H \). To make life easy consider the simplest case where this basis is also an eigenbasis for the second operator \( B \) for all but two of that operators eigenvectors. For such a system let’s write

to be zero. Because of (15) only the matrix elements with respect to states \( \ket{1}, \ket{2} \) need be considered. Two of the matrix elements above are clearly zero, regardless of the values of \( B_{1 1}\), and \(B_{2 2} \), and for the other two to be zero, we must either have

and \( B \ket{2} = B_{2 2} \ket{2} \). This contradicts the requirement that \( \ket{1}, \ket{2} \) not be eigenkets of \( B \), leaving only the second option. That second option means there must be a degeneracy in the system.